A question about approximations. We use approximations because they give a neat and clean answer and they are quite useful in physical situations too. For example, we can write: $\frac {5} {1000 + 0.15}$ as $\frac{5} {1000}$ , because both fractions are different only by $0.000000074988$, which is quite small if we are making some calculations for physical situations.
I'm having a doubt about when we can and when we can't just remove the very small term which is in addition or subtraction. For instance, it is a general thing to do $$ \frac {a} {b+c} = \frac {a} {b} \text{ if } b\gg c$$ 
Now, let me explain my doubt through an example of physics:   
Electric field due to a continuous line of charge of length $2L$ at a point $z$ above it's midpoint is given by ($\lambda$ the  being line charge density) $$ \frac{1}{4\pi\epsilon_0}~~\frac {2\lambda L} {z \sqrt{z^2 +L^2} } $$ now if $z\gt\gt L$ then we have , $$ \frac{1}{4\pi\epsilon_0} ~~ \frac {2\lambda L } {z^2 } $$ considering $L^2 = 0$ and this is quite consistent with the intuition that if we are too far away from the line it would look to us like a point charge with the total charge of $\lambda ~2L$.   
Electric field due to continuous surface charge of circular disk of radius R (with surface charge density $\sigma$) at a point $z$ above it's center is given by  :- $$ \frac {1} {2\epsilon_0} ~~ \sigma z ~~\left[ \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2} }\right]$$  now if $z\gt\gt R$ and if we mechanically remove $R^2 $ by considering it too small we would get $0$ which is quite non-intuitive because if we go far far away from that disk we will see that all the charge is concentrated at the center and the formula should become $$ \frac{1} {4 \pi\epsilon_0} ~~ \frac  {\sigma \pi R^2 ~z}{z^2} $$ but we won't get it if we put $R^2 = 0$ . My question is why we can't do this here but can do it above? 
 A: We don't just put $L=0$ in the first equation: We take the first (non-zero) term from the Taylor expansion. For your second formula, you can use the generalised Binomial theorem:
$$(1+x)^\alpha=\sum_{n=0}^{\infty} \binom{\alpha}{n} x^n$$
Which means that
$$(1+x)^{-1/2} \approx 1-\frac{1}{2}x$$
HINT:
$$\frac{1}{\sqrt{z^2+R^2}}=\frac{1}{\sqrt{z^2(1+R^2/z^2)}}=\frac{1}{z}\frac{1}{\sqrt{1+R^2/z^2}}=\frac{1}{z}(1+R^2/z^2)^{-1/2}$$
A: Similar to Botond's answer.
Considering 
$$y=z \left( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2} }\right)=1- \frac{z}{\sqrt{R^2 + z^2}}=1-\sqrt{\frac{z^2}{{R^2 + z^2}} }=1-\sqrt{1-\frac{R^2}{{R^2 + z^2}} }$$
Now, using long division or Taylor series for large values of $z$
$$\frac{R^2}{R^2 + z^2}=\frac{R^2}{z^2}-\frac{R^4}{z^4}+O\left(\frac{1}{z^6}\right)$$
$$\sqrt{1-\frac{R^2}{{R^2 + z^2}} }=\sqrt{1-\frac{R^2}{z^2}+\frac{R^4}{z^4}+O\left(\frac{1}{z^6}\right)}=1-\frac{R^2}{2 z^2}+\frac{3 R^4}{8 z^4}+O\left(\frac{1}{z^6}\right)$$ which makes
$$y=\frac{R^2}{2 z^2}-\frac{3 R^4}{8 z^4}+O\left(\frac{1}{z^6}\right)=\frac{R^2}{2 z^2}+O\left(\frac{1}{z^4}\right)$$
